This document provides an overview of additional aspects of acid-base equilibria, including: 1. Important relations for pH, pOH, Ka, and Kb calculations. 2. Steps for calculating pH, pKa, [H+], Ka for acids or bases, including distinguishing between strong and weak acids/bases. 3. Steps for calculating pH for mixtures of acids and bases, including considerations for salt solutions, buffer solutions, and calculating pH changes upon adding small amounts of acid or base to a buffer.

1. CM4106 Chemical Equilibria & Thermodynamics
Lesson 3 (Part 1)
Additional Aspects of Acid-Base Equilibria
(Topic 3.1 – 3.5)
A Chemistry Education Blog by Mr Tan
http://chemistry-mr-tan-yong-yao.blogspot.sg/

2. Fundamentals:
Important relations
p = -log10
At 25ºC:
pH + pOH = pKw = 14
pKa + pKb = 14
[H+][OH–] = 10-14
Ka x Kb = Kw = 10-14

3. (I) Calculations for acids OR bases
pH, pKa, [H+], Ka pOH, pKb, [OH–], Kb
Step 1: Determine what is present in the solution.
(A) Acid : Strong Acid vs Weak Acid
Monoprotic Acid / Diprotic Acid / Triprotic Acid
Concentration of Acid
(B) Base: Strong Base vs Weak Base
Monoprotic base / Diprotic base / Triprotic base
Concentration of Bas
Take note of concentration of acid / bases
(For dilute solutions, we need to take into consideration of
[H+] / [OH-] from auto-ionization from water)
Step 2: Use the appropriate equations for the respective species.

4. (I) pH calculations of Acid OR Base
x2
Ka =
Strong Acid Weak Acid ([HA] – x)
Strong acids dissociate HA(aq) ⇌ H+(aq) + A-(aq)
completely into ions in
aqueous solution.
I [HA] 0 0
C -x +x +x
[H+] = [HA] E [HA] - x +x +x
x2
Kb =
Strong Base Weak Base ([B] – x)
Strong bases dissociate B + H2O ⇌ BH+ + OH-
completely into ions in
I [B] - 0 0
aqueous solution.
C -x - +x +x
[OH-] = [B] E [B] - x - x x

5. (I) pH calculations of Acid OR Base
1. Determine if acid/ base is strong or weak (more common)
2. For weak acids, if asked to determine Ka, pH or [H+], you can save time by
using the formula:
[H+] = Ka × c [OH–] = Kb × c
Assumption: x is negligible
Example:
Calculate the pH of 0.50M HF solution at 25ºC where the Ka = 7.1 x 10-4
[H+][F-] x2
Ka = = = 7.1 x 10-4
[HF] (0.50 – x)
Assumption: For weak acids, x must be very small  0.50 – x ≈ 0.5
x2
= 7.1 x 10-4 x = [H+] = 0.0188 M Assumption is valid;
0.50
x < 5% of [HF]initial
pH = 1.73 (to 2 d.p.)

6. Calculate the pH of the following solutions at 298K
0.10 mol dm−3 CH3COOH (pKa = 4.75) Weak acid solution
(monoprotic acid)
CH3COOH(aq) + H2O (l) ⇌ CH3COO(aq) + H3O+(aq)
Initial (M) 0.10 - 0.00 0.00
Change (M) -x - +x +x
Eqm (M) 0.10 - x - +x +x
[CH3COO][H3O+] x2
Ka = = = 10 4.75 = 1.79 x 10-5
[CH3COOH] (0.10 – x)
Assumption: For weak acids, x must be very small  0.10 – x ≈ 0.10
x2 Remember to validate Assumption
= 1.79 x 10-5 Assumption is justified,
0.10
x < 5% of [CH3COOH]initial
x = [H3O+] = 1.338 x 10-3 M
Concentration: 2 s.f.
pH = 2.87 ( 2 d.p.) pH: 2 d.p.

7. (II) Calculations for mixture of acids AND bases
pH, pKa, [H+], α pOH, pKb, [OH–]
Step 1: Determine what is present in the solution.
(A) acid Stoichiometric amounts of acid and base –
base salt solution
(B) buffer Non-stoichiometric amounts of acid and
base – likely to be a buffer
(C) salt
Step 2: Use the appropriate equations for the respective
species.

8. (II) (a) or (c) pH calculation of salt solutions
COVERED IN GREATER DETAILS IN TOPIC 3.8
Salt solutions can be (i) neutral (ii) weak acids or (iii) weak bases
Basic Salt
CH3COONa (aq) → CH3COO- (aq) + Na+ (aq)
CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)
conjugate base pH > 7
Acidic Salt
NH4Cl (aq) → NH4+ (aq) + Cl- (aq)
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
conjugate acid pH < 7

9. (II) (b) pH of Buffer Solutions
Acid Buffer weak acid + conjugate base
weak acid Assume negligible dissociation of acid due to common ion effect
CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)
CH3COONa(aq) → CH3COO-(aq) + Na+(aq)
conjugate base
Assume full dissociation of salt [salt]
pH = pKa + lg
(strong electrolyte) [acid]
Basic Buffer weak base + conjugate acid
weak base Assume negligible dissociation of base due to common ion effect
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
NH4Cl(aq) → NH4+(aq) + Cl-(aq)
conjugate acid
Assume full dissociation of salt
(strong electrolyte) pOH = pKb + lg [salt]
[base]

10. (II) (b) pH of Buffer Solutions
1. Determine if buffer is acidic/ basic
2. When using H-H equation, you can save time by just
calculating no. of moles of salt and acid/base since total
volume is the same and cancels out. Useful for MCQs!
nsalt/V nsalt/V
pH = pKa + lg pOH = pKb + lg
nacid/V nbase/V
3. Need to be sensitive to the condition (maximum
buffering capacity)
where [salt] = [acid] which simply means pH = pKa
Similarly for basic buffer,
[salt] = [base]  pOH = pKb

11. [salt]
pH  pK a  lg
Sample Calculation [acid]
• Find the pH of a buffer made from adding 3.28 g of sodium
ethanoate, CH3COO-Na+ to 1.00 dm3 of 0.0100 M ethanoic acid,
CH3COOH . (Ka = 1.80 x 10-5)
Useful Information
Molar Mass of CH3COO-Na+ = 82.03 g/mol
Amount of CH3COO-Na+ present in 3.28 g = (3.28 g / 82.03)
= 0.0400 mol
[CH3COO-Na+] = 0.0400 M
[CH3COOH] = 0.0100 M
[salt]
pH  pK a  lg
[acid]
5  0.0400 
  lg(1.80  10 )  lg   pH = 5.3466
 0.0100  = 5.347 (3 d.p.)
Page 46

12. An alternative approach –
Sample Calculation ICE Table
• Find the pH of a buffer made from adding 3.28 g of sodium ethanoate,
CH3COO-Na+ to 1.00 dm3 of 0.0100 M ethanoic acid, CH3COOH.
(Ka = 1.80 x 10-5)
n(CH3COO-Na+) present in 3.28 g = (3.28 g / 82.03)
Useful Information = 0.0400 mol
Molar Mass of CH3COO -Na+ = 82.03 g/mol
[CH3COO-Na+] = 0.0400 M
[CH3COOH] = 0.0100 M
CH3COOH(aq) + H2O (l) ⇌ H3O+ (aq) + CH3COO (aq)
Initial [ ] (M) 0.0100 M - 0 0.0400 M
Change [ ] (M) -x - +x +x
Equilibrium [ ] (M) 0.0100 - x - x 0.0400 + x
Ka =
[H3O+][CH3COO-] x (0.0400 + x)
= = 1.80 x 10-5
[CH3COOH] (0.0100 – x)
Solve for x, pH = - lg [H3O+] = - lg (4.497 x 10-6)
x = [H3O+] = 4.497 x 10-6 M
= 5.3470
= 5.347 (3 d.p.)

13. Calculating pH change after small amounts of
acid / base is added to buffer solutions
1) Determine new [salt]new and [acid]new / [base]new
How?
a) Determine n(acid / base) added to buffer solution
b) Determine the change to n(salt)buffer and n(acid)buffer /
n(base)buffer by considering the neutralization action of a
buffer
c) Determine new total volume of the buffer solution
[salt]new
pH new  pK a  lg
[acid] new

14. CM4106
Chemical Equilibria &Thermodynamics
Review of Pre-Quiz 3
Topic 3 (3.1 – 3.5)
Additional Aspects of Acid-Base Equilibria
Buffer Solutions, pH calculation, Maximum Buffer Capacity

15. Question 1 (a)
• Hypochlorous acid, HOCl, is a weak acid commonly
used as a bleaching agent. It dissociates in water as
represented by the equation below.
HOCl(aq) + H2O(l) ⇌ OCl-(aq) + H3O+(aq)
Ka = 3.2 x10-8
a) Write the equilibrium-constant expression for the
dissociation of HOCl in water.
[OC l  ][H 3O  ]
Ka 
[HOC l ]

16. Question 1(b)
b) Calculate the molar concentration of H3O+ in a
0.14 M solution of HOCl.
HOCl(aq) + H2O(l) ⇌ OCl-(aq) + H3O+(aq)
Ka = 3.2 x10-8
HOCl (aq) + H2O (l) ⇌ H3O+ (aq) + OCl- (aq)
Initial [ ] (M) 0.14 - 0 0
Change [ ] (M) -y - +y +y
Equilibrium [ ] (M) 0.14 – x - x y
y2 Solve for y,
Ka 
(0.14  y)
y = 6.7 x 10-5
y2
3.2  10 8  [H3O+] = 6.7 x 10-5 M (2 s.f.)
(0.14  y )
Note: units must be present

17. Question 1 (c)
c) A mixture of HOCl and sodium hypochlorite
(NaOCl) can be used as a buffer.
Write two equations to show how this buffer solution
can control pH.
• Addition of H3O+:
OCl(aq) + H3O+(aq)  HOCl(aq) + H2O(l)
Note: state symbols must be present
• Addition ofOH :
HOCl(aq) + OH(aq)  OCl (aq) + H2O(l)

18. Question 1(d)
HOCl reacts with NaOH according to the reaction represented
below:
HOCl(aq) + OH-(aq)  OCl-(aq) + H2O(l)
Bob, a budding young chemist, decides to make a buffer by
adding a volume of 10.0 mL of 0.56 M NaOH to 50.0 mL of 0.14 M
HOCl solution. Assume that the volumes are additive.
HOCl(aq) + H2O(l) ⇌ OCl-(aq) + H3O+(aq)
Ka = 3.2 x10-8
Preliminary Considerations:
Reaction of NaOH with HOCl will form NaOCl (salt containing conjugate
base OCl-).
Hence, if HOCl is in excess, the final mixture will contain some excess
HOCl that is unreacted and some NaOCl formed as a result of the
neutralization reaction.  BUFFER SOLUTION (Mixture of acid and
conjugagte base (salt)

19. Question 1(d)(i)
• Calculate the pH of the buffer solution
Henderson-Hasselbalch Equation to
[salt]
pH  pK a  log calculate pH of buffer equation
[acid] Need to determine [salt] / [acid]
HOCl (aq) + NaOH (aq)  NaOCl (aq) + H2O(l)
n HOCl = (50.0/1000)(0.14) = 7.0 x 10-3 mol (excess reagent)
n NaOH = (10.0/1000)(0.56) = 5.6 x 10-3 mol
Total Volume = 50.0 + 10.0 = 60.0 cm3 = 0.0600 L
n HOCl remaining n NaOCl formed = 5.6 x10-3 mol
= 7.0 x 10-3 - 5.6 x 10-3
= 1.4 x 10-3 mol
[OCl] = 5.6 x 10-3 mol / 0.0600 L
[HOCl] = 1.4 x 10-3 mol / 0.0600 L = 9.33 x 10-2 M
= 2.33 x 10-2 M

20. Question 1(d)(i)
• Calculate the pH of the buffer solution.
[HOCl] = 1.4 x 10-3 mol / 0.0600 L [OCl] = 5.6 x 10-3 mol / 0.0600 L
= 2.33 x 10-2 M = 9.33 x 10-2 M
[salt]
pH  pK a  log
[acid]
8 (9.33  10 -2 )
pH   lg(3.2 10 )  lg
(2.33  10 -2 )
pH  8.097

21. Question 1(d)(ii)
• Bob would like to prepare a new buffer solution. How many
grams of solid NaOH must he add to a 50.0 mL of 0.20 M HOCl
to obtain a buffer that has a pH of 6.0?
• Assume that the addition of the solid NaOH results in a
negligible change in volume. (Mr NaOH = 40)
Let the no of mole of NaOH required be y mol:
n HOCl = (50/1000)(0.20) = 0.01 mol n HOCl remaining = (0.01 - y) mol
n NaOH = y mol n NaOCl formed = y mol
NaOH (s) + HOCl (aq)  NaOCl (aq) + H2O (l)

22. Question 1(d)(ii)
• Bob would like to prepare a new buffer solution. How many
grams of solid NaOH must he add to a 50.0 mL of 0.20 M HOCl
to obtain a buffer that has a pH of 6.0?
• Assume that the addition of the solid NaOH results in a
negligible change in volume. (Mr NaOH = 40)
n HOCl remaining = (0.01 - y) mol Total Volume = 50.0 mL = 0.0500 L
n NaOCl formed = y mol
 y 
 
[salt]
6.0  - log (3.2 x 10 )  lg  0.0500  y
pH  pK a  log
-8
 1.4949  lg
[acid]  0.01  y  0.01  y
 
 0.0500 
Solve for y,
Mass of NaOH (s) required y
= 3.101 x 10-4 x 40 y = n (NaOCl) formed 10 1.4949 
= n (NaOH) required
0.01  y
= 0.1240 g
= 3.101x 10-4 mol